Minimum Operations to Reduce X to Zero - Video Solutions

Problem Overview: You are given an array nums and an integer x. In one operation you remove the leftmost or rightmost element from the array and subtract its value from x. The goal is to reduce x to exactly zero using the minimum number of operations.

The key observation: removing elements from both ends is equivalent to keeping a contiguous middle subarray. If the total sum of the array is total, the removed elements must sum to x, meaning the remaining subarray must sum to total - x. The task becomes finding the longest subarray with that sum.

Approach 1: Sliding Window for the Complement (O(n) time, O(1) space)

This is the optimal approach and the one most interviewers expect. First compute target = totalSum - x. Instead of thinking about removing elements, search for the longest subarray whose sum equals target. Use a classic sliding window with two pointers. Expand the right pointer while accumulating the window sum, and shrink the left pointer whenever the sum exceeds the target. When the window sum equals the target, record the maximum length. If the longest valid subarray length is L, the minimum operations equals n - L. This works because all values are positive, allowing the window to move monotonically.

Approach 2: Prefix Sum + Hash Table (O(n) time, O(n) space)

Another way to find the longest subarray with sum target uses prefix sums and a hash table. Maintain a running prefix sum while iterating through the array. Store the earliest index where each prefix sum appears. If the current prefix sum is curr, check whether curr - target exists in the map. If it does, the elements between those indices form a valid subarray with the desired sum. Track the maximum length seen. This approach works for more general arrays but uses extra memory for the hash map.

Approach 3: Dynamic Programming (O(n * x) time, O(x) space)

A dynamic programming formulation treats the problem as choosing elements from the left or right to reach sum x. Define dp[s] as the minimum operations needed to reach sum s. Iteratively update states as you consider elements from both ends. While this approach demonstrates the problem’s combinational nature, it becomes inefficient when x is large. For typical constraints, the sliding window strategy is significantly faster and simpler.

Recommended for interviews: The sliding window complement trick is the expected solution. It reduces the original problem to a longest subarray search in O(n) time and constant space. Mentioning the prefix-sum alternative shows broader understanding, while the sliding window demonstrates strong problem transformation skills.